This example is describing the Chromatography method.
At the complete combustion of butane the carbon dioxide and water are formed.
2C₄H₁₀ + 13O₂ = 8CO₂ + 10H₂O
C₄H₁₀ ⇒ 4CO₂, 5H₂O ⇒ 13O ⇒ 6.5O₂
we use integer coefficients: 8CO₂, 10H₂O, 13O₂
Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Answer:

Explanation:
1. Moles of CCl₄

2. Molar mass of CCl₄
MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol
3. Mass of CCl₄

4. Volume of CCl₄
